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Let $f:E\subset\mathbb{R^n}\rightarrow\mathbb{R^n}$ be a $C^1$ function, where $E$ is open. Show that $f$ is locally Lipschitz-continuous, that is, for all compact $K\subset E$ we have $f$ restrict to $K$ Lipschitz-continuous.

I try to take a open ball in $E$ to force convexity, and use a auxiliary function $g: [0,1] \rightarrow \mathbb{R^n}$ such that $g(t) = f((1-t)x+ty)$, but this don't work.

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marked as duplicate by John B, JonMark Perry, Community May 29 '16 at 23:39

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I'll do the $\mathbb{R}$ case. The reasoning can be generalized to $\mathbb{R}^n$. Let $x\in E$ and $\epsilon>0$ small enough, then as $f$ is $C^1$, there exists a neighborhood of $x$ (we can take a sphere $B(x,r)$ of some radius $r>0$) such that $$f'(x')\in[f'(x)-\epsilon,f'(x)+\epsilon].$$ Therefore, for any two points $y,z\in B(x,r)$ we have $$|f(y)-f(z)| = \left|\int_y^zf'(s)ds\right|\le|z-y|(|f'(x)|+\epsilon)$$ (for the general case, integrate over the straight line relying $y$ and $z$).

Now cover your compact by a finite number of such balls.

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  • $\begingroup$ Technically I don't know integral yet. Do you know another way? Thanks for help $\endgroup$ – Renan R. May 28 '16 at 1:28
  • $\begingroup$ @RenanR. Take a look at the answer mentioned in the comment of Jonas in your OP, then. $\endgroup$ – Daniel Robert-Nicoud May 28 '16 at 11:06

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